\(n_{Al}=\dfrac{3,78}{27}=0,14\left(mol\right)\\ Al+XCl_3\rightarrow AlCl_3+X\\ m_{giảm}=4,06\left(g\right)=m_X-m_{Al}\\n_X=n_{XCl_3}=n_{AlCl_3}=n_{Al}=0,14\left(mol\right)\\ m_X=m_{Al}+4,06=3,78+4,06=7,84\left(g\right)\\ M_X=\dfrac{7,84}{0,14}=56\left(\dfrac{g}{mol}\right)\\ Vậy.XCl_3.là:FeCl_3 \)